Question

What are the solutions to
log3x + log3(x2 + 2) = 1 + 2log3x?

x = –2

x = –1

x = 1

x = 2

There is no true solution.

Answer 1

If this was a MC test and you know how to take the log_3 of numbers, I say just guess and check

Answer 2

ok... just a quick tip... you can't take the log of a negative number...

\[\log_{3}(-1) ~~doesn't ~exist\]

so you can eliminate A and B

Answer 3

If not, since this is a question with log_3, we have to take 3 and raisee it to each expression since \[\Large a^{\log_{a}x} =x\]

Answer 4

so then you are left with using the log laws

the 1st law you need

\[\log(a) + \log(b) = \log(ab)\]

next you need to start the right hand side

\[\log_{a}(a) = 1\]

also on the right you need

\[a \log(x) = \log(x^a)\]

so if you use the log laws... you can gte an equation to solve

Answer 5

thanks guys i figured it out, it was C and D

Answer 6

\[\Large \log_{3}x + \log_{3}(x^2 + 2) = 1 + 2\log_{3}x\]
\[\Huge 3^{\log_{3}x + \log_{3}(x^2 + 2) = 1 + 2\log_{3}x}\]

Answer 7

\[\Large x+(x^2+2)=3+x^2\]

Answer 8

x=1

Answer 9

well that's not quite right

if you apply the log laws

it becomes

\[\log_{3})x^3 + 2x) = \log_{3}(3x^2)\]

now you can equate things

\[x^3 + 2x = 3x^2\]

and subsequently you can solve for x.

Answer 10

which becomes

\[x^3 - 3x^2 + 2x = 0~~~or~~~x(x^2 - 3x + 2) = 0\]

factor the quadratic and you can find all the solutions

but you should ignore x = 0 as a solution as it doesn't exist

Answer 11

Wait, where did you get an x^3 from?

Answer 12

If I'm not mistaken, these are log_3s that we are dealing with, and my math is correct then going with that, if the first erm is just an x term, the next is an x^2 which evaluated at x=1 gives me 1. and the other side is just 1+log_3(1^2) which is just 1 as well @campbell_st